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Consider a permutation matrix $P$ and two vectors $x$, $v$ with 2-norm = 1 and all positive entries.

Are the optimal solutions $P^\ast$ of $\max_P \; (x^T \cdot Py)$ and $\min_P \; \|x-Py\|_2$ the same?

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  • $\begingroup$ Minimizing $\|x-Py\|^2_2=\|x\|^2_2+\|y\|^2-2(x^T\cdot Py)$ over $P$ is equivalent to maximizing $x^T\cdot Py$. $\endgroup$ – Did Oct 26 '11 at 6:57
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Hint:

$$ \|x-Py\|_2^2=\|x\|_2^2+\|Py\|_2^2-2x^T\cdot Py=2(1-x^T\cdot Py) $$

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  • $\begingroup$ Thanks!! One more question, if both of the vectors are able to permute freely, is the best strategy equals to sort both vectors in either ascending or descending order? $\endgroup$ – Benjamin Oct 26 '11 at 7:05
  • $\begingroup$ @Benjamin That approach works. In fact, you can fix one of the permutations arbitrarily, and optimize over the second permutation. (Note that $\| Px - Qy \| = \| x - P^{-1} Q y \|$.) $\endgroup$ – Srivatsan Oct 26 '11 at 7:09
  • $\begingroup$ @SrivatsanNarayanan Thanks:) $\endgroup$ – Benjamin Oct 26 '11 at 7:15
  • $\begingroup$ @Benjamin That is the statement of the Rearrangement inequality. (It's not neccessary to assume x, y are unit norm and positive, by the way.) $\endgroup$ – p.s. Nov 9 '11 at 21:34

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